2019高考数学二轮复专题数列课件练
2019高考数学二轮复专题数列课件练（1）
等差数列等数列基问题
1等差数列{an}前9项等前4项a11ak+a40k
2已知等数列{an}中a32a4a616(a_7 a_9)(a_3 a_5 )
3(2018江苏南通中学高三考前刺练)已知等差数列{an}公差d3Sn前n项a1a2a9成等数列S5值
4(2018南通高三第二次调研)设等数列{an}前n项SnS3S9S6成等差数列a83a5
5设数列{an}首项a11满足a2n+12a2n1a2na2n1+1数列{an}前20项
6(2018江苏锡常镇四市高三教学情况调研(二))已知公差d等差数列{an}前n项SnS_10S_5 4(4a_1)d
7已知Sn数列{an}前n项a12〖〖S_n〗_+〗_12Sn设bnlog2an1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )值
8(2018扬州高三第三次调研)已知实数abc成等数列a+6b+2c+1成等差数列b值
9(2018扬州高三第三次调研)已知数列{an}满足an+1+(1)nan(n+5)2(n∈N*)数列{an}前n项Sn
(1)求a1+a3值
(2)a1+a52a3
①求证数列{a2n}等差数列
②求满足S2p4S2m(pm∈N*)数(pm)
10(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}首项1公差d数列{bn}前n项Sn意n∈N*6Sn9bnan2恒成立
(1)果数列{Sn}等差数列证明数列{bn}等差数列
(2)果数列{b_n+12}等数列求d值
(3)果d3数列{cn}首项1cnbnbn1(n≥2)证明数列{an}中存穷项表示数列{cn}中两项
答案精解精析
1答案 10
解析 S9S49a1+36d4a1+6da1+6da70a4+a102a70k10
2答案 4
解析 等数列中奇数项符号相a3>0a5>0a4a6〖a_5〗^216a54a78a916(a_7 a_9)(a_3 a_5 )( 8)( 2)4
3答案 652
解析 题意a1a9〖a_2〗^2a1(a1+24)(a1+3)2解a112S55×12+(5×4)2×3652
4答案 6
解析 S3S9S6成等差数列S3+S62S9等数列{an}公q1时成立q≠1(a_1 ( 1 q^3 ) )(1 q)+(a_1 ( 1 q^6 ) )(1 q)2(a_1 ( 1 q^9 ) )(1 q)化简2q6q310q312(舍1)
a5a_8q^3 6
5答案 2056
解析 题意奇数项构成等数列a1+a3+…+a19(1 2^10)(1 2)1023偶数项a2+a4+…+a20(a1+1)+(a3+1)+…+(a19+1)1033数列{an}前20项2056
6答案 2
解析 〖S_1〗_0S_5 4〖S_1〗_04S510a1+45d4(5a1+10d)(4a_1)d2
7答案 1910
解析 〖〖S_n〗_+〗_12SnS1a12数列{Sn}首项公2等数列Sn2nn≥2时anSnSn12n2n12n1a12适合an{■(2 n1 @2^(n 1) n≥2 )┤bn{■(1 n1 @n 1 n≥2 )┤1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )1+1(1×2)+1(2×3)+…+1(9×10)1+(1 12)+(12 13)+…+(19 110)21101910
8答案 34
解析 设等数列abc公q(q≠0)abqcbqa+6bq+6b+2c+1bq+1成等差数列(bq+6)+(bq+1)2(b+2)化简b32 (q+1q) b时q<0时
q+1q≤2b32 (q+1q) ≤34仅q1时取等号b值34
9解析 (1)条件{■(a_2 a_13① @a_3+a_272 ② )┤
②①a1+a312
(2)①证明an+1+(1)nan(n+5)2
{■(a_2n a_(2n 1)(2n+4)2 ③ @a_(2n+1)+a_2n(2n+5)2 ④ )┤
④③a2n1+a2n+112
112+12(a1+a3)+(a3+a5)4a3
a314(1)知a1+a312a114
a2n114(a_(2n 3) 14)…
(1)n1(a_1 14)0
a2n114代入③式
a2nn+94
a2(n+1)a2n1(常数)数列{a2n}等差数列
②易知a1a2n+1S2na1+a2+…+a2n
(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
n^22+3n
S2p4S2m知p^22+3p4(m^22+3m)
(2m+6)2(p+3)2+27
(2m+p+9)(2mp+3)27pm∈N*
2m+p+9≥122m+p+92mp+3均正整数{■(2m+p+927 @2m p+31 )┤解p10m4
求数(104)
10解析 (1)证明设数列{Sn}公差d'
∵6Sn9bnan2①
6Sn19bn1an12(n≥2)②
①②6(SnSn1)9(bnbn1)(anan1)③
6d'9(bnbn1)dbnbn1(6d' +d)9(n≥2)常数
{bn}等差数列
(2)③6bn9bn9bn1d3bn9bn1+d
{b_n+12}等数列(b_n+12)(b_(n 1)+12)(3b_(n 1)+d3+12)(b_(n 1)+12)(3(b_(n 1)+12)+d3 1)(b_(n 1)+12)3+(d3 1)(b_(n 1)+12)(n≥2)n关常数
d310bn1+12常数
d310时d3符合题意
bn1+12常数时
6Sn9bnan2中令n16b19b1a12a11解b11
bn1+12b1+1232(n≥2)
时3+(d3 1)(b_(n 1)+12)3+(d3 1)(32)1
解d6
综d3d6
(3)证明d3时an3n2
(2)数列{b_n+12}32首项3公等数列bn+1232×3n1123nbn12(3n1)
n≥2时cnbnbn112(3n1)12(3n11)3n1
n1时满足式
cn3n1(n≥1)
设anci+cj(1≤i果i≥23n3倍数3i1+3j13倍数3n(3i1+3j1)3倍数
23倍数矛盾
i13n3+3j1n1+3j2(j234…)
数列{an}中存穷项表示数列{cn}中两项
2019高考数学二轮复专题数列课件练（2）
等差数列等数列基问题
1等差数列{an}前9项等前4项a11ak+a40k
2已知等数列{an}中a32a4a616(a_7 a_9)(a_3 a_5 )
3(2018江苏南通中学高三考前刺练)已知等差数列{an}公差d3Sn前n项a1a2a9成等数列S5值
4(2018南通高三第二次调研)设等数列{an}前n项SnS3S9S6成等差数列a83a5
5设数列{an}首项a11满足a2n+12a2n1a2na2n1+1数列{an}前20项
6(2018江苏锡常镇四市高三教学情况调研(二))已知公差d等差数列{an}前n项SnS_10S_5 4
(4a_1)d
7已知Sn数列{an}前n项a12〖〖S_n〗_+〗_12Sn设bnlog2an1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )值
8(2018扬州高三第三次调研)已知实数abc成等数列a+6b+2c+1成等差数列b值
9(2018扬州高三第三次调研)已知数列{an}满足an+1+(1)nan(n+5)2(n∈N*)数列{an}前n项Sn
(1)求a1+a3值
(2)a1+a52a3
①求证数列{a2n}等差数列
②求满足S2p4S2m(pm∈N*)数(pm)
10(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}首项1公差d数列{bn}前n项Sn意n∈N*6Sn9bnan2恒成立
(1)果数列{Sn}等差数列证明数列{bn}等差数列
(2)果数列{b_n+12}等数列求d值
(3)果d3数列{cn}首项1cnbnbn1(n≥2)证明数列{an}中存穷项表示数列{cn}中两项
答案精解精析
1答案 10
解析 S9S49a1+36d4a1+6da1+6da70a4+a102a70k10
2答案 4
解析 等数列中奇数项符号相a3>0a5>0a4a6〖a_5〗^216a54a78a916(a_7 a_9)(a_3 a_5 )( 8)( 2)4
3答案 652
解析 题意a1a9〖a_2〗^2a1(a1+24)(a1+3)2解a112S55×12+(5×4)2×3652
4答案 6
解析 S3S9S6成等差数列S3+S62S9等数列{an}公q1时成立q≠1(a_1 ( 1 q^3 ) )(1 q)+(a_1 ( 1 q^6 ) )(1 q)2(a_1 ( 1 q^9 ) )(1 q)化简2q6q310q312(舍1)a5a_8q^3 6
5答案 2056
解析 题意奇数项构成等数列a1+a3+…+a19(1 2^10)(1 2)1023偶数项a2+a4+…+a20(a1+1)+(a3+1)+…+(a19+1)1033数列{an}前20项2056
6
答案 2
解析 〖S_1〗_0S_5 4〖S_1〗_04S510a1+45d4(5a1+10d)(4a_1)d2
7答案 1910
解析 〖〖S_n〗_+〗_12SnS1a12数列{Sn}首项公2等数列Sn2nn≥2时anSnSn12n2n12n1a12适合an{■(2 n1 @2^(n 1) n≥2 )┤bn{■(1 n1 @n 1 n≥2 )┤1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )1+1(1×2)+1(2×3)+…+1(9×10)1+(1 12)+(12 13)+…+(19 110)21101910
8答案 34
解析 设等数列abc公q(q≠0)abqcbqa+6bq+6b+2c+1bq+1成等差数列(bq+6)+(bq+1)2(b+2)化简b32 (q+1q) b时q<0时q+1q≤2b32 (q+1q) ≤34仅q1时取等号b值34
9解析 (1)条件{■(a_2 a_13① @a_3+a_272 ② )┤
②①a1+a312
(2)①证明an+1+(1)nan(n+5)2
{■(a_2n a_(2n 1)(2n+4)2 ③ @a_(2n+1)+a_2n(2n+5)2 ④ )┤
④③a2n1+a2n+112
112+12(a1+a3)+(a3+a5)4a3
a314(1)知a1+a312a114
a2n114(a_(2n 3) 14)…
(1)n1(a_1 14)0
a2n114代入③式
a2nn+94
a2(n+1)a2n1(常数)数列{a2n}等差数列
②易知a1a2n+1S2na1+a2+…+a2n
(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
n^22+3n
S2p4S2m知p^22+3p4(m^22+3m)
(2m+6)2(p+3)2+27
(2m+p+9)(2mp+3)27pm∈N*
2m+p+9≥122m+p+92mp+3均正整数{■(2m+p+927 @2m p+31 )┤解p10m4
求数(104)
10解析 (1)证明设数列{Sn}公差d'
∵6Sn9bnan2①
6Sn19bn1an12(n≥2)②
①②6(SnSn1)9(bnbn1)(anan1)③
6d'9(bnbn1)dbnbn1(6d' +d)9(n≥2)常数
{bn}等差数列
(2)③6bn9bn9bn1d3bn9bn1+d
{b_n+12}等数列(b_n+12)(b_(n 1)+12)(3b_(n 1)+d3+12)(b_(n 1)+12)(3(b_(n 1)+12)+d3 1)(b_(n 1)+12)3+(d3 1)(b_(n 1)+12)(n≥2)n关常数
d310bn1+12常数
d310时d3符合题意
bn1+12常数时
6Sn9bnan2中令n16b19b1a12a11解b11
bn1+12b1+1232(n≥2)
时3+(d3 1)(b_(n 1)+12)3+(d3 1)(32)1
解d6
综d3d6
(3)证明d3时an3n2
(2)数列{b_n+12}32首项3公等数列bn+1232×3n1123nbn12(3n1)
n≥2时cnbnbn112(3n1)12(3n11)3n1
n1时满足式
cn3n1(n≥1)
设anci+cj(1≤i果i≥23n3倍数3i1+3j13倍数3n(3i1+3j1)3倍数
23倍数矛盾
i13n3+3j1n1+3j2(j234…)
数列{an}中存穷项表示数列{cn}中两项
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2019高考数学二轮复专题数列课件练（1）
等差数列等数列基问题
1等差数列{an}前9项等前4项a11ak+a40k
2已知等数列{an}中a32a4a616(a_7 a_9)(a_3 a_5 )
3(2018江苏南通中学高三考前刺练)已知等差数列{an}公差d3Sn前n项a1a2a9成等数列S5值
4(2018南通高三第二次调研)设等数列{an}前n项SnS3S9S6成等差数列a83a5
5设数列{an}首项a11满足a2n+12a2n1a2na2n1+1数列{an}前20项
6(2018江苏锡常镇四市高三教学情况调研(二))已知公差d等差数列{an}前n项SnS_10S_5 4(4a_1)d
7已知Sn数列{an}前n项a12〖〖S_n〗_+〗_12Sn设bnlog2an1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )值
8(2018扬州高三第三次调研)已知实数abc成等数列a+6b+2c+1成等差数列b值
9(2018扬州高三第三次调研)已知数列{an}满足an+1+(1)nan(n+5)2(n∈N*)数列{an}前n项Sn
(1)求a1+a3值
(2)a1+a52a3
①求证数列{a2n}等差数列
②求满足S2p4S2m(pm∈N*)数(pm)
10(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}首项1公差d数列{bn}前n项Sn意n∈N*6Sn9bnan2恒成立
(1)果数列{Sn}等差数列证明数列{bn}等差数列
(2)果数列{b_n+12}等数列求d值
(3)果d3数列{cn}首项1cnbnbn1(n≥2)证明数列{an}中存穷项表示数列{cn}中两项
答案精解精析
1答案 10
解析 S9S49a1+36d4a1+6da1+6da70a4+a102a70k10
2答案 4
解析 等数列中奇数项符号相a3>0a5>0a4a6〖a_5〗^216a54a78a916(a_7 a_9)(a_3 a_5 )( 8)( 2)4
3答案 652
解析 题意a1a9〖a_2〗^2a1(a1+24)(a1+3)2解a112S55×12+(5×4)2×3652
4答案 6
解析 S3S9S6成等差数列S3+S62S9等数列{an}公q1时成立q≠1(a_1 ( 1 q^3 ) )(1 q)+(a_1 ( 1 q^6 ) )(1 q)2(a_1 ( 1 q^9 ) )(1 q)化简2q6q310q312(舍1)
a5a_8q^3 6
5答案 2056
解析 题意奇数项构成等数列a1+a3+…+a19(1 2^10)(1 2)1023偶数项a2+a4+…+a20(a1+1)+(a3+1)+…+(a19+1)1033数列{an}前20项2056
6答案 2
解析 〖S_1〗_0S_5 4〖S_1〗_04S510a1+45d4(5a1+10d)(4a_1)d2
7答案 1910
解析 〖〖S_n〗_+〗_12SnS1a12数列{Sn}首项公2等数列Sn2nn≥2时anSnSn12n2n12n1a12适合an{■(2 n1 @2^(n 1) n≥2 )┤bn{■(1 n1 @n 1 n≥2 )┤1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )1+1(1×2)+1(2×3)+…+1(9×10)1+(1 12)+(12 13)+…+(19 110)21101910
8答案 34
解析 设等数列abc公q(q≠0)abqcbqa+6bq+6b+2c+1bq+1成等差数列(bq+6)+(bq+1)2(b+2)化简b32 (q+1q) b时q<0时
q+1q≤2b32 (q+1q) ≤34仅q1时取等号b值34
9解析 (1)条件{■(a_2 a_13① @a_3+a_272 ② )┤
②①a1+a312
(2)①证明an+1+(1)nan(n+5)2
{■(a_2n a_(2n 1)(2n+4)2 ③ @a_(2n+1)+a_2n(2n+5)2 ④ )┤
④③a2n1+a2n+112
112+12(a1+a3)+(a3+a5)4a3
a314(1)知a1+a312a114
a2n114(a_(2n 3) 14)…
(1)n1(a_1 14)0
a2n114代入③式
a2nn+94
a2(n+1)a2n1(常数)数列{a2n}等差数列
②易知a1a2n+1S2na1+a2+…+a2n
(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
n^22+3n
S2p4S2m知p^22+3p4(m^22+3m)
(2m+6)2(p+3)2+27
(2m+p+9)(2mp+3)27pm∈N*
2m+p+9≥122m+p+92mp+3均正整数{■(2m+p+927 @2m p+31 )┤解p10m4
求数(104)
10解析 (1)证明设数列{Sn}公差d'
∵6Sn9bnan2①
6Sn19bn1an12(n≥2)②
①②6(SnSn1)9(bnbn1)(anan1)③
6d'9(bnbn1)dbnbn1(6d' +d)9(n≥2)常数
{bn}等差数列
(2)③6bn9bn9bn1d3bn9bn1+d
{b_n+12}等数列(b_n+12)(b_(n 1)+12)(3b_(n 1)+d3+12)(b_(n 1)+12)(3(b_(n 1)+12)+d3 1)(b_(n 1)+12)3+(d3 1)(b_(n 1)+12)(n≥2)n关常数
d310bn1+12常数
d310时d3符合题意
bn1+12常数时
6Sn9bnan2中令n16b19b1a12a11解b11
bn1+12b1+1232(n≥2)
时3+(d3 1)(b_(n 1)+12)3+(d3 1)(32)1
解d6
综d3d6
(3)证明d3时an3n2
(2)数列{b_n+12}32首项3公等数列bn+1232×3n1123nbn12(3n1)
n≥2时cnbnbn112(3n1)12(3n11)3n1
n1时满足式
cn3n1(n≥1)
设anci+cj(1≤i
23倍数矛盾
i13n3+3j1n1+3j2(j234…)
数列{an}中存穷项表示数列{cn}中两项
2019高考数学二轮复专题数列课件练（2）
等差数列等数列基问题
1等差数列{an}前9项等前4项a11ak+a40k
2已知等数列{an}中a32a4a616(a_7 a_9)(a_3 a_5 )
3(2018江苏南通中学高三考前刺练)已知等差数列{an}公差d3Sn前n项a1a2a9成等数列S5值
4(2018南通高三第二次调研)设等数列{an}前n项SnS3S9S6成等差数列a83a5
5设数列{an}首项a11满足a2n+12a2n1a2na2n1+1数列{an}前20项
6(2018江苏锡常镇四市高三教学情况调研(二))已知公差d等差数列{an}前n项SnS_10S_5 4
(4a_1)d
7已知Sn数列{an}前n项a12〖〖S_n〗_+〗_12Sn设bnlog2an1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )值
8(2018扬州高三第三次调研)已知实数abc成等数列a+6b+2c+1成等差数列b值
9(2018扬州高三第三次调研)已知数列{an}满足an+1+(1)nan(n+5)2(n∈N*)数列{an}前n项Sn
(1)求a1+a3值
(2)a1+a52a3
①求证数列{a2n}等差数列
②求满足S2p4S2m(pm∈N*)数(pm)
10(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}首项1公差d数列{bn}前n项Sn意n∈N*6Sn9bnan2恒成立
(1)果数列{Sn}等差数列证明数列{bn}等差数列
(2)果数列{b_n+12}等数列求d值
(3)果d3数列{cn}首项1cnbnbn1(n≥2)证明数列{an}中存穷项表示数列{cn}中两项
答案精解精析
1答案 10
解析 S9S49a1+36d4a1+6da1+6da70a4+a102a70k10
2答案 4
解析 等数列中奇数项符号相a3>0a5>0a4a6〖a_5〗^216a54a78a916(a_7 a_9)(a_3 a_5 )( 8)( 2)4
3答案 652
解析 题意a1a9〖a_2〗^2a1(a1+24)(a1+3)2解a112S55×12+(5×4)2×3652
4答案 6
解析 S3S9S6成等差数列S3+S62S9等数列{an}公q1时成立q≠1(a_1 ( 1 q^3 ) )(1 q)+(a_1 ( 1 q^6 ) )(1 q)2(a_1 ( 1 q^9 ) )(1 q)化简2q6q310q312(舍1)a5a_8q^3 6
5答案 2056
解析 题意奇数项构成等数列a1+a3+…+a19(1 2^10)(1 2)1023偶数项a2+a4+…+a20(a1+1)+(a3+1)+…+(a19+1)1033数列{an}前20项2056
6
答案 2
解析 〖S_1〗_0S_5 4〖S_1〗_04S510a1+45d4(5a1+10d)(4a_1)d2
7答案 1910
解析 〖〖S_n〗_+〗_12SnS1a12数列{Sn}首项公2等数列Sn2nn≥2时anSnSn12n2n12n1a12适合an{■(2 n1 @2^(n 1) n≥2 )┤bn{■(1 n1 @n 1 n≥2 )┤1(b_1 b_2 )+1(b_2 b_3 )+…+1(b_10 b_11 )1+1(1×2)+1(2×3)+…+1(9×10)1+(1 12)+(12 13)+…+(19 110)21101910
8答案 34
解析 设等数列abc公q(q≠0)abqcbqa+6bq+6b+2c+1bq+1成等差数列(bq+6)+(bq+1)2(b+2)化简b32 (q+1q) b时q<0时q+1q≤2b32 (q+1q) ≤34仅q1时取等号b值34
9解析 (1)条件{■(a_2 a_13① @a_3+a_272 ② )┤
②①a1+a312
(2)①证明an+1+(1)nan(n+5)2
{■(a_2n a_(2n 1)(2n+4)2 ③ @a_(2n+1)+a_2n(2n+5)2 ④ )┤
④③a2n1+a2n+112
112+12(a1+a3)+(a3+a5)4a3
a314(1)知a1+a312a114
a2n114(a_(2n 3) 14)…
(1)n1(a_1 14)0
a2n114代入③式
a2nn+94
a2(n+1)a2n1(常数)数列{a2n}等差数列
②易知a1a2n+1S2na1+a2+…+a2n
(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
n^22+3n
S2p4S2m知p^22+3p4(m^22+3m)
(2m+6)2(p+3)2+27
(2m+p+9)(2mp+3)27pm∈N*
2m+p+9≥122m+p+92mp+3均正整数{■(2m+p+927 @2m p+31 )┤解p10m4
求数(104)
10解析 (1)证明设数列{Sn}公差d'
∵6Sn9bnan2①
6Sn19bn1an12(n≥2)②
①②6(SnSn1)9(bnbn1)(anan1)③
6d'9(bnbn1)dbnbn1(6d' +d)9(n≥2)常数
{bn}等差数列
(2)③6bn9bn9bn1d3bn9bn1+d
{b_n+12}等数列(b_n+12)(b_(n 1)+12)(3b_(n 1)+d3+12)(b_(n 1)+12)(3(b_(n 1)+12)+d3 1)(b_(n 1)+12)3+(d3 1)(b_(n 1)+12)(n≥2)n关常数
d310bn1+12常数
d310时d3符合题意
bn1+12常数时
6Sn9bnan2中令n16b19b1a12a11解b11
bn1+12b1+1232(n≥2)
时3+(d3 1)(b_(n 1)+12)3+(d3 1)(32)1
解d6
综d3d6
(3)证明d3时an3n2
(2)数列{b_n+12}32首项3公等数列bn+1232×3n1123nbn12(3n1)
n≥2时cnbnbn112(3n1)12(3n11)3n1
n1时满足式
cn3n1(n≥1)
设anci+cj(1≤i
23倍数矛盾
i13n3+3j1n1+3j2(j234…)
数列{an}中存穷项表示数列{cn}中两项
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